An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the level ground 150m downrange from the launch point throws an apple straight up with the minimum intial speed necessary to meet the path of the arrow.
(a) what is the initial speed of the apple?
(b) at what time after the arrow launch should the apple be thrown so that the arrow hits the apple?

Respuesta :

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = [tex]\dfrac{150}{45 cos 50^0}[/tex]

t  =  5.19 s

a) height of arrow

[tex]s = u t +\dfrac{1}{2}gt^2[/tex]

[tex]s = v sin \theta \times t+\dfrac{1}{2}gt^2[/tex]

[tex]s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2[/tex]

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = [tex]\dfrac{u}{g}=\dfrac{30.29}{9.81}[/tex]

                                             = 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s

The initial speed of the apple will be 30.29 m/s.

How to calculate the speed

The following can be deduced from the information given:

Velocity = 45m/s

Angle = 50°

Horizontal velocity = 45cos50° = 28.93°

Time = 150/28.93 = 5.19s

The initial speed will be calculated thus:

From the second law of motion,

s = ut + 1/2gt²

where, g = acceleration due to gravity = 9.81

s = (45sin50° × 5.19) - (1/2 × 9.81 × 5.19²)

s = (45 × 0.766 × 5.19) - 132.12

s = 178.90 - 132.12

s = 46.78m

Since v² - u² = 2gs

Making u the subject of the formula goes thus:

u² = 2 × 9.81 × 46.78

u² = 917.82

u = 30.29m/s

Furthermore, the time taken by the apple will be:

= u/g = 30.29/9.81 = 3.09s

The time after the arrow launch that the apple should be thrown will be:

= 5.19s - 3.09s

= 2.1 seconds.

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